Question 1196798
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<pre>
The square of the modulus of this number,  {{{sqrt(2)/2}}} + {{{(sqrt(2)/2)*i}}}, is

    {{{r^2}}} = {{{(sqrt(2)/2)^2}}} + {{{(sqrt(2)/2)^2}}} = {{{2/4 + 2/4}}} = {{{1/2}}} + {{{1/2}}} = 1.


So, the modulus itself is  r = {{{sqrt(1)}}} = 1.


The argument of this number, "a", satisfies 

    tan(a) = {{{((sqrt(2)/2))/((sqrt(2)/2))}}} = 1.


So, the argument is  a = {{{pi/4}}} = 45°.


Hence, the square of this number has the modulus {{{1^2}}} = 1 and the argument 2a = {{{pi/2}}} = 90°.


It means that the square of the given number is the complex imaginary unit "i".
</pre>

At this point, the proof is complete.